my problem is that i'm using two voltage dividers in a circuit ..... both r getting effected by each other.... is there any way that i could isolate them from each other..... i've explained the matter in a better way in the image.....

the frequency of the square wave is 300bps.....

plz help....and keep it as simple as possible and briefly explain the solution too..... thanks

problem 2:
at some other place i've a 300bps square wave which has 2.5V as lo and 3V as hi..... i want to change it to a 0-5 square wave..... how would i achieve this???? i thought of achieving it in two steps... first bring the lo voltage back to 0V, like remove the DC component and then amplify it to 5V using a single supply op amp...... plz comment if there is a better solution.....

OK, on the first part of the problem you have to remember that only the AC component of the square wave comes through. This would be a 5VP-P, or 2.5P waveform. On the other side of the cap you would have a square wave that goes from 1.5V to 6.5. Your voltage divider would drop that to 20.4% (1KΩ / (1KΩ + 3.9KΩ)), so that square wave would become .5VP. 4VDC + .5 is 4.5V, 4VDC - .5V is 3.5V. This is where your wave form is coming from.

Frankly, I can't make heads or tails from your second drawing. Are you connecting the 250Ω resistor to the chips +5V line? If so you need to state this clearer, don't assume people will know what you are trying to show. Being able to draw a clear schematic is crucial to electronics communication.

As far as your 2nd problem, a comparator will make for an easy solution.
R1 is a potentiometer that sets the "trip" level.
R2 is a pull-up resistor for the open-collector comparator output.
R3 provides hysteresis which helps prevent false triggering.

the schematic Sargent suggested worked perfect...... the signal was a a pure 0-5V output.... the simulation shows some noise in the upper level bt that wasnt observed in the real hardware..... thanks Sargent.... now could u plz answer my following questions, how did u find the values of the Resistors and how did u set the triggering point ? like is there any equation with which u can set the triggering point. if yes plz tell me for future use...... thanks

@ Bill
thanks Bill, for the solution...... really really appreciate it..... will try in the lab and pot the results.....

the schematic Sargent suggested worked perfect...... the signal was a a pure 0-5V output.... the simulation shows some noise in the upper level bt that wasnt observed in the real hardware..... thanks Sargent.... now could u plz answer my following questions, how did u find the values of the Resistors and how did u set the triggering point ? like is there any equation with which u can set the triggering point. if yes plz tell me for future use...... thanks

@ Bill
thanks Bill, for the solution...... really really appreciate it..... will try in the lab and pot the results.....

the schematic Sargent suggested worked perfect...... the signal was a a pure 0-5V output..

Click to expand...

Good to know you got it working. However, if you look at the output signal again, you should note that it does not go all the way to ground; the lowest the voltage should be going is around 90mV. This is due to the limitations of how much current the output transistor of the comparator can sink, and the resulting voltage drop.

.. the simulation shows some noise in the upper level bt that wasnt observed in the real hardware...

Click to expand...

That was expected. The simulation has no capacitance or inductance added. A "real" circuit will have at least a small amount of parasitic capacitance. Even a few pF would be enough to squelch the noise.

.. thanks Sargent.... now could u plz answer my following questions, how did u find the values of the Resistors and how did u set the triggering point?

Click to expand...

To minimize current draw, I simply selected a value for R1 that would result in less than 1mA current from 5v to ground. 5v across 10k Ohms = 0.5mA current.
As far as R1's wiper setting, your stated signal voltage was 2.5v to 3v. Since the supply voltage is 5v, 2.5v/5v=50%, 3v/5v=60%. Since in the simulation the pot was installed upside-down, the wiper was set at 45% instead of 55%. This set the threshold level at 5v x 55% = 2.75v.
Different values of resistance could certainly be used, but current levels greater than 1mA would be excessive IMHO, and current levels less than 0.1mA would make the threshold much more susceptible to noise.

R2 was selected to provide roughly 3.33mA current for the comparators' output to sink; the target was between 3mA and 3.5mA. The comparators' output can sink somewhat more current than that, but as it sinks more current, the minimum output voltage increases.

R3 was chosen to arbitrarily provide roughly 1/10 the current flow of R1, roughly 125mV of hysteresis, or 1/4 of your input signal level's peak to peak voltage. 3V-2.5V=0.5V; 0.5V/4=125mV. Actual hysteresis is somewhat less due to R1's offset from center, but it should still work OK.